Question

solve for x. round to the nearest tenth if necessary. * rectangle with height 16, base 33, and diagonal x (right triangle formed by height, base, and diagonal) options: x = 49; x = 43.2; x = 36.7; x = 28.9; x = 18.3

Explanation:

Step1: Identify the triangle type

The figure is a rectangle, so the triangle formed by the diagonal is a right triangle with legs 16 and 33, and hypotenuse \( x \). We use the Pythagorean theorem: \( a^2 + b^2 = c^2 \), where \( a = 16 \), \( b = 33 \), and \( c = x \).

Step2: Apply the Pythagorean theorem

Substitute the values into the formula: \( x^2 = 16^2 + 33^2 \). Calculate \( 16^2 = 256 \) and \( 33^2 = 1089 \). Then \( x^2 = 256 + 1089 = 1345 \).

Step3: Solve for \( x \)

Take the square root of 1345: \( x = \sqrt{1345} \approx 36.7 \). Wait, no, wait: Wait, 16 squared is 256, 33 squared is 1089. 256 + 1089 is 1345? Wait, no, 256 + 1089: 256 + 1000 is 1256, plus 89 is 1345. Then square root of 1345: let's calculate. 36 squared is 1296, 37 squared is 1369. So 1345 is between 36² and 37². 36.7 squared: 36.7 36.7. Let's calculate 3636=1296, 360.7=25.2, 0.736=25.2, 0.70.7=0.49. So (36 + 0.7)² = 36² + 236*0.7 + 0.7² = 1296 + 50.4 + 0.49 = 1346.89. Oh, that's very close to 1345. So \( \sqrt{1345} \approx 36.7 \) (since 36.7² ≈ 1346.89, which is a bit more than 1345, but maybe my calculation was off. Wait, no, wait: 16 and 33. Wait, maybe I made a mistake. Wait, 16 squared is 256, 33 squared is 1089. 256 + 1089 = 1345. Then square root of 1345: let's use a calculator. \( \sqrt{1345} \approx 36.7 \). Wait, but let's check the options. One of the options is \( x = 36.7 \). So that's the answer.

Wait, but wait, maybe I messed up the legs. Wait, the rectangle has length 33 and width 16, so the diagonal is the hypotenuse. So Pythagorean theorem: \( x = \sqrt{16^2 + 33^2} = \sqrt{256 + 1089} = \sqrt{1345} \approx 36.7 \). So the correct option is \( x = 36.7 \).

Answer:

\( \boldsymbol{x = 36.7} \) (corresponding to the option "x = 36.7")